3sqrt{3y-1}+sqrt[3]{1-2x}=0 eching

$$3 \sqrt { 3 y - 1 } + \sqrt[ 3 ] { 1 - 2 x } = 0$$

y uchun yechish

$$y=\frac{\left(1-2x\right)^{\frac{2}{3}}+9}{27}$$

$$x\geq \frac{1}{2}$$

Ildizli tenglamalarni yechish qadamlari

Tenglamaning ikkala tarafidan \sqrt[3]{1-2x} ni ayirish.

$$3\sqrt{3y-1}+\sqrt[3]{1-2x}-\sqrt[3]{1-2x}=-\sqrt[3]{1-2x}$$

O‘zidan \sqrt[3]{1-2x} ayirilsa 0 qoladi.

$$3\sqrt{3y-1}=-\sqrt[3]{1-2x}$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3\sqrt{3y-1}}{3}=-\frac{\sqrt[3]{1-2x}}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$\sqrt{3y-1}=-\frac{\sqrt[3]{1-2x}}{3}$$

Tenglamaning ikkala taraf kvadratini chiqarish.

$$3y-1=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}$$

1 ni tenglamaning ikkala tarafiga qo'shish.

$$3y-1-\left(-1\right)=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}-\left(-1\right)$$

O‘zidan -1 ayirilsa 0 qoladi.

$$3y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}-\left(-1\right)$$

\frac{\left(1-2x\right)^{\frac{2}{3}}}{9} dan -1 ni ayirish.

$$3y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1$$

Ikki tarafini 3 ga bo‘ling.

$$\frac{3y}{3}=\frac{\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1}{3}$$

3 ga bo'lish 3 ga ko'paytirishni bekor qiladi.

$$y=\frac{\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1}{3}$$

\frac{\left(1-2x\right)^{\frac{2}{3}}}{9}+1 ni 3 ga bo'lish.

$$y=\frac{\left(1-2x\right)^{\frac{2}{3}}}{27}+\frac{1}{3}$$

x uchun yechish (complex solution)

$$x=\frac{81\sqrt{3y-1}y-27\sqrt{3y-1}+1}{2}$$

$$|\frac{arg(3y-1)}{2}-arg(-\sqrt[3]{-\left(3y-1\right)^{\frac{3}{2}}})|<\frac{2\pi }{3}\text{ or }y=\frac{1}{3}$$

y uchun yechish (complex solution)

$$y = \frac{1}{3} = 0,3333333333333333$$

$$x = \frac{1}{2} = 0,5$$

x uchun yechish

$$x=\frac{81\sqrt{3y-1}y-27\sqrt{3y-1}+1}{2}$$

$$y\geq \frac{1}{3}$$